## Pieri and Giambelli Formulas

It’s been a few weeks, but now I’m back and today we’ll talk about the multiplication in the cohomology ring of Grassmannians.  Though we won’t talk about the Littlewood-Richardson rule in its full glory, we will howver discuss the special cases of the Pieri rule and the Giambelli formula.

Last time, we saw that the cohomology of a Grassmannian had a basis parameterized by Young diagrams.  The real powerhouse is the Pieri rule, for one simple reason: it completely determines the multiplication.  Once it’s been determined, we can steal formulas from anything else that happens to use the Pieri rule, and there are good candidates, which we’ll discuss later in the post.  For now, here’s the Pieri rule:

Pieri Rule: Let $\lambda$ be any Young diagram and let $k$ be the Young diagram with only one row of boxes of length $k$.  Then $\sigma_\lambda \sigma_k=\sum \sigma_\nu$ where $\nu$ runs over the the Young diagrams obtained from $\lambda$ by adding $k$ boxes, no two in the same column.

We note quickly that diagrams that don’t fit into the box given by the Grassmannian we’re working on just give Schubert class zero, and that with this the rule is perfectly fine without reference to any specific Grassmannian.  We’ll make a bit of use of this later.  For now, we’ll prove this:

Proof: To prove this, the first thing we need is the dual Young diagram.  Given a Young diagram $\lambda$ inside of an $m\times n$ rectangle, we define the dual, $\lambda^*$, to be the diagram obtained by deleting $\lambda$ from the rectangle, and then rotating 180 degrees.  This is an extremely useful notion in the Schubert calculus, for the following reason:

Claim: Let $\lambda,\mu$ be two Young diagrams contained in an $m\times n$ rectangle such that $|\lambda|+|\mu|=mn$.  Then $\sigma_\lambda\cup \sigma_\mu=\delta_{\mu,\lambda^*}$ in the top cohomology group.

This claim can be seen by choosing a basis and writing out matrices as we did a few posts ago, and noting that the dual can be represented by the matrix with ones in teh same positions, but with the zeros and stars switched, so the two spaces must intersect at a unique point.

Now, to show this, we introduce the opposite flag, $\tilde{F}$, whose $k$th term is given by the span of the last $k$ basis vectors for our vector space, and write the Schubert variety found using that flag as $\tilde{\Omega}_\lambda$.    (Note, working the matrices out for these carefully is what proves the above claim)

Now, set $A_i=F_{n+i-\lambda_i}$, $B_i=\tilde{F}_{n+i-\mu_i}$ and $C_i=A_i\cap B_{r+1-i}$ for given $\lambda,\mu$.  These are useful when working with the intersection of $\Omega_\lambda$ and $\tilde{\Omega}_\mu$.

Now, to show Pieri’s formula, we just need to show that both sides have the same intersection numbers with all $\mu$ which have $|\mu|=rn-|\lambda|-k$ in our Grassmannian.  So that means that we’d need $\sigma_\mu\sigma_\lambda\sigma_k=1$ whenever we have that $\mu\subset \lambda^*$, and that the boxes in $\lambda^*$ but not in $\mu$ are all in different columns.  This is just the condition $n-\lambda_r\geq \mu_1\geq n-\lambda_{r-1}\geq\mu_2\geq\ldots\geq n-\lambda_1\geq \mu_r\geq 0$.

So now we take $\Omega_\lambda,\tilde{\Omega}_\mu$ and define $\Omega_k(L)$ by taking $L$ to be a general linear subspace of dimension $n+1-k$.  Now, we’ve just used the word “general” and we’re going to talk much more about it next time.  The idea is that on the Grassmannian of these things, there might be a few $n+1-k$ planes that need to be avoided, but they form subvarieties of positive codimension, so we have an open, dense set to work with.  Pretty much we can choose a plane at random, and it will, with probability 1, be ok.

Now, let $C$ be the span of the $C_i$, and set $A_0=B_0=0$.  Then, we have that $C=\cap_{i=0}^r (A_i+B_{r-i})$, $\sum_{i=1}^r \dim(C_i)=r+k$ and that $C=C_1+\ldots+C_r$ is a direct sum iff we have the situation with the Young diagrams of $\lambda$ and $\mu$ as above.  Now, if that condition fails, the triple intersection $\Omega_\lambda\cap \tilde{\Omega}_\mu\cap \Omega_k(L)=\emptyset$.  We just need to show that it’s a single point if it does hold.

Now, any point in $\Omega_\lambda\cap \tilde{\Omega}_\mu$ will have $\dim(V\cap A_i)\geq i$ and $\dim(V\cap B_{r+1-i})\geq r+1-i$, by definition, and so we will get $\dim(V\cap C_i)\geq i+(r+1-i)-r=1$.  So now, if the $C_i$ are linearly independent, then $V\supset \oplus (V\cap C_i)$, which has big enough dimension that they are equal.  So $V$ is a direct sum of $V\cap C_i$, one for each dimension of $V$, and none of them being trivial.  So all must have dimension 1.

This tells us that $C=\oplus C_i$, and a generic $L$ will meet it in a line $\mathbb{C}\cdot v$, with $v=u_1\oplus\ldots\oplus u_r$ with $u_i\in C_i$.  Now, $V\subset C$, and so must contain $v$, and so $u_i\in V$, so $V$ is the subspace spanned by the $u_i$, and is the unique point in which the three Schubert varieties intersect, giving us the Pieri rule.  $\Box$.

The upshot from our point of view is that this lets us start doing some computations.  Look at $Gr(2,4)$, and say we want to compute $\sigma_1\sigma_1$.  The Pieri rule says that this will be $\sigma_2+\sigma_{1,1}$, very quickly.  Next week, we’ll use this fact to start solving actual problems.  In the meantime, there are other formulas that are worth mentioning, as well as connections to things that I understand a bit less well.

The big theoretical thing the Pieri Rule does for us is that it tells us that there’s a surjective ring homomorphism $\Lambda\to H^*(Gr^n(\mathbb{C}^m),\mathbb{Z})$, where $\Lambda$ is the ring of symmetric functions.  The map is given by taking the Schur function $s_\lambda$ to the cohomology class $\sigma_\lambda$.  Now, I know very little about Schur functions, so I’m not even going to bother defining them.  The upshot is that there are lots of formulas known in $\Lambda$, and we can just use them in our cohomology calculations.  One of the better consequences is the following:

Giambelli Formula: $\sigma_\lambda=\det(\sigma_{\lambda_i+j-i})_{1\leq i,j\leq r}$

This formula says that we can represent any Schubert class in terms of the “special” Schubert classes that the Pieri rule tells us how to multiply.  So with these two formulas we can, in principle, perform any multiplication we could possibly desire in the cohomology ring of the Grassmannian.  And example of the Giambelli formula in $Gr(2,4)$ is that we can write $\sigma_{1,1}=\det\left(\begin{array}{cc}\sigma_1&\sigma_2\\ \sigma_0&\sigma_1\end{array}\right)=\sigma_1^2-\sigma_2$.  This formula isn’t terribly exciting because we already obtained it by using Pieri on $\sigma_1^2$, but Giambelli is useful in more difficult situations as well.

All this culminates in the fact that there are numbers called the Littlewood-Richardson coefficients, $c_{\lambda\mu}^\nu$, which are defined completely combinatorially (there may be a guest post on this) which are the structure constants.  That is, they’re positive integers such that $\sigma_\lambda\sigma_\mu=\sum_\nu c_{\lambda\mu}^\nu \sigma_\nu$.